Black and Red balls in a box

Let the number of red balls be $x$ and the number of black balls be $y$ $\begin{aligned}p&=\dfrac{\dbinom{x}{2}}{\dbinom{x+y}{2}}\\\\ q&=\dfrac{\dbinom{y}{2}}{\dbinom{x+y}{2}}\\\\ p-q&=\dfrac{\dbinom{x}{2}-\dbinom{y}{2}}{\dbinom{x+y}{2}}=\dfrac{x(x-1)-y(y-1)}{(x+y)(x+y-1)}\\\\ &=\dfrac{(x^2-y^2)-(x-y)}{(x+y)(x+y-1)}=\dfrac{(x-y)(x+y-1)}{(x+y)(x+y-1)}\\\\ &=\dfrac{x-y}{x+y}\hspace{7mm}\color{#3D99F6}\small x+y\neq1 \text{ ,As we need at least 2 balls of each color for p and q to be nonzero}\\\\ \implies \dfrac{x-y}{x+y}&=\dfrac{23}{37}\hspace{7mm}\color{#3D99F6}\small p-q=\dfrac{23}{37}\\\\ (x-y)&=\dfrac{23}{37}\cdot(x+y)\\\\ \text{Since } x \text{ and } y &\text{ are positive integers and }x+y<1000 \text{ we have }\\\\ (x-y)&\leq 23\cdot \left \lfloor{\dfrac{1000}{37}}\right \rfloor\\\\ \implies(x-y)&\leq 23\times27\\\\ \implies (x-y)&\leq \color{#EC7300}\boxed{\color{#333333}621}\end{aligned}$

Let the number of red balls be r, the number of black balls be b and the total number of balls be n.

Then:

$r= n - b$

$p = \frac { n - b }{n} × \frac { (n - b) - 1 }{ n- 1}$

$q = \frac { b }{n} × \frac { b - 1 }{ n- 1 }$

$p - q = \frac { (n - b) ((n - b) - 1) - b(b - 1) }{ n(n- 1) } = \frac { n^2 -2bn+b^2-n+b-b^2+b }{ n(n- 1) } =$

$= \frac { n^2 - n - 2bn + 2b }{ n(n- 1) } =\frac { (n - 2b)(n-1) }{ n(n- 1) } = \frac { n-2b}{n} = 1- \frac { 2b}{n}$

Now:

$1- \frac { 2b }{ n } = \frac { 23 }{ 37 }$

$\frac { b }{ n } = \frac { 7 }{ 37 } \Rightarrow b = \frac { 7 }{ 37 } n \Rightarrow r = \frac { 30 }{ 37 } n$

$D = r - b = \frac { 23 }{ 37 } n$

Since n < 1000 and b and r are integers :

$D_ { \max } = \frac { 23 }{ 37 } n_ { \max } = \frac { 23 }{ 37 } × \lfloor \frac { 1000 }{ 37 } \rfloor × 37= \frac { 23 }{ 37 } × 999 = \boxed { 621 }$

We can also find that:

$n_ { \max } = 999 \Rightarrow r_ { \max } = 810 , b_ { \max } = 189$